Integration by Substitution

The basic rules handle sums of powers, exponentials, and trig functions. But what about $\int 2x(x^2 + 3)^5\,dx$ or $\int x^2\cos(x^3)\,dx$? These came from the chain rule, and substitution (often called u-substitution) is the chain rule run backwards. It’s the single most useful integration technique in a first calculus course (AP Calculus AB or IB).

Key ideas

Where it comes from

The chain rule says $\dfrac{d}{dx}F\big(g(x)\big) = F'\big(g(x)\big)\,g'(x)$. Reading that backwards:

$$\int f\big(g(x)\big)\,g'(x)\,dx = F\big(g(x)\big) + C, \quad \text{where } F' = f.$$

Look for an “inside function” $g(x)$ whose derivative $g'(x)$ also appears as a factor.

The steps

1. Choose $u$: usually the inside of a power, a root, an exponential, a trig function, or a denominator.
2. Find $du$: $du = u'(x)\,dx$.
3. Rewrite the whole integral in terms of $u$ and $du$. No $x$ may be left over.
4. Integrate with respect to $u$.
5. Substitute back so the answer is in terms of $x$.

Fixing constants

If $du$ is off by a constant factor, that’s fine: solve for $dx$ or multiply and divide by the constant. For example, with $u = x^3$, $du = 3x^2\,dx$, so $x^2\,dx = \tfrac{1}{3}\,du$.

You can only fix constants this way. If a variable is missing, a different $u$ or a different method is needed.

Definite integrals: change the limits

For a definite integral, you can change the limits to $u$-values when you substitute:

$$\int_a^b f\big(g(x)\big)\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du .$$

Then evaluate in terms of $u$ and never go back to $x$. (Or substitute back to $x$ and use the original limits. Just don’t mix the two.)

Choosing a technique

When you meet a new integral, ask in this order:

1. Is it a basic rule? Use the table from indefinite integrals.
2. Can algebra make it basic? Expand, split a fraction with a one-term denominator, or rewrite roots as powers.
3. Is there an inside function whose derivative is (nearly) there? Use substitution.
4. Is it a rational function with a polynomial on top that’s too big, or a quadratic on the bottom? Use long division or completing the square.

Worked examples

Example 1: A perfect match

Find $\displaystyle\int 2x(x^2 + 3)^5\,dx$.

Solution. The inside of the power is $x^2 + 3$, and its derivative $2x$ is right there. Let $u = x^2 + 3$, so $du = 2x\,dx$:

$$\int 2x(x^2 + 3)^5\,dx = \int u^5\,du = \frac{u^6}{6} + C = \frac{(x^2 + 3)^6}{6} + C$$

Check: $\dfrac{d}{dx}\dfrac{(x^2 + 3)^6}{6} = \dfrac{6(x^2 + 3)^5 \cdot 2x}{6} = 2x(x^2 + 3)^5$. ✓

Example 2: Adjusting a constant

Find (a) $\displaystyle\int x^2\cos(x^3)\,dx$ and (b) $\displaystyle\int e^{5x}\,dx$.

Solution.

(a) Let $u = x^3$, so $du = 3x^2\,dx$ and $x^2\,dx = \tfrac{1}{3}\,du$:

$$\int x^2\cos(x^3)\,dx = \frac{1}{3}\int \cos u\,du = \frac{1}{3}\sin u + C = \frac{1}{3}\sin(x^3) + C$$

(b) Let $u = 5x$, so $dx = \tfrac{1}{5}\,du$:

$$\int e^{5x}\,dx = \frac{1}{5}\int e^u\,du = \frac{1}{5}e^{5x} + C$$

A handy shortcut: for a linear inside function $ax + b$, just divide by $a$. For example, $\int \cos(4x)\,dx = \tfrac{1}{4}\sin(4x) + C$.

Example 3: Getting a logarithm

Find (a) $\displaystyle\int \frac{x}{x^2 + 1}\,dx$ and (b) $\displaystyle\int \tan x\,dx$.

Solution.

(a) Let $u = x^2 + 1$, so $du = 2x\,dx$ and $x\,dx = \tfrac{1}{2}\,du$:

$$\int \frac{x}{x^2 + 1}\,dx = \frac{1}{2}\int \frac{1}{u}\,du = \frac{1}{2}\ln|u| + C = \frac{1}{2}\ln(x^2 + 1) + C$$

(No absolute value is needed at the end because $x^2 + 1$ is always positive.)

(b) Write $\tan x = \dfrac{\sin x}{\cos x}$. Let $u = \cos x$, so $du = -\sin x\,dx$:

$$\int \frac{\sin x}{\cos x}\,dx = -\int \frac{1}{u}\,du = -\ln|u| + C = -\ln|\cos x| + C$$

Whenever the numerator is (a constant times) the derivative of the denominator, the answer is a logarithm.

Example 4: A definite integral, changing the limits

Evaluate $\displaystyle\int_0^2 \frac{x}{\sqrt{x^2 + 5}}\,dx$.

Solution. Let $u = x^2 + 5$, so $x\,dx = \tfrac{1}{2}\,du$. Change the limits:

• when $x = 0$, $u = 5$;
• when $x = 2$, $u = 9$.

$$\begin{aligned} \int_0^2 \frac{x}{\sqrt{x^2 + 5}}\,dx &= \frac{1}{2}\int_5^9 u^{-1/2}\,du \\ &= \frac{1}{2}\Big[2u^{1/2}\Big]_5^9 \\ &= \Big[\sqrt{u}\Big]_5^9 = 3 - \sqrt{5} \end{aligned}$$

Common mistakes

Leaving some x behind. After substituting, the integral must be entirely in $u$. If you have $\int u^5 \cdot x\,du$, the substitution isn’t finished (or it’s the wrong $u$).

Dropping the du (or the dx). $du = 2x\,dx$ tells you what to replace. Writing it out keeps your constants right.

Fixing a missing variable as if it were a constant. In $\int \cos(x^3)\,dx$, the $x^2$ from $du = 3x^2\,dx$ is missing. You can’t just divide by $3x^2$. (In fact, this integral has no elementary antiderivative.)

Using the old limits with u. In Example 4, writing $\tfrac{1}{2}\int_0^2 u^{-1/2}\,du$ gives the wrong answer. Either change both limits to $u$-values, or go back to $x$ before substituting $0$ and $2$.

Forgetting to substitute back. For an indefinite integral, the final answer must be in terms of $x$.

Practice

1. (Warm-up) Find $\displaystyle\int 3x^2(x^3 - 1)^4\,dx$.

Solution

Let $u = x^3 - 1$, $du = 3x^2\,dx$:

$$\int u^4\,du = \frac{u^5}{5} + C = \frac{(x^3 - 1)^5}{5} + C$$

2. (Warm-up) Find $\displaystyle\int \cos(4x)\,dx$.

Solution

Let $u = 4x$, $dx = \tfrac{1}{4}\,du$:

$$\frac{1}{4}\int \cos u\,du = \frac{1}{4}\sin(4x) + C$$

3. (Warm-up) Find $\displaystyle\int 2x\,e^{x^2}\,dx$.

Solution

Let $u = x^2$, $du = 2x\,dx$:

$$\int e^u\,du = e^{x^2} + C$$

4. (Core) Find $\displaystyle\int \frac{x}{(x^2 + 4)^3}\,dx$.

Solution

Let $u = x^2 + 4$, so $x\,dx = \tfrac{1}{2}\,du$:

$$\frac{1}{2}\int u^{-3}\,du = \frac{1}{2} \cdot \frac{u^{-2}}{-2} + C = -\frac{1}{4(x^2 + 4)^2} + C$$

5. (Core) Find $\displaystyle\int \frac{(\ln x)^2}{x}\,dx$.

Solution

Let $u = \ln x$, $du = \tfrac{1}{x}\,dx$:

$$\int u^2\,du = \frac{u^3}{3} + C = \frac{(\ln x)^3}{3} + C$$

6. (Core) Find $\displaystyle\int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx$.

Solution

Let $u = \sqrt{x} = x^{1/2}$, so $du = \dfrac{1}{2\sqrt{x}}\,dx$ and $\dfrac{dx}{\sqrt{x}} = 2\,du$:

$$2\int e^u\,du = 2e^{\sqrt{x}} + C$$

7. (Core) Evaluate $\displaystyle\int_0^1 x(x^2 + 1)^3\,dx$.

Solution

Let $u = x^2 + 1$, $x\,dx = \tfrac{1}{2}\,du$. When $x = 0$, $u = 1$; when $x = 1$, $u = 2$.

$$\frac{1}{2}\int_1^2 u^3\,du = \frac{1}{2}\left[\frac{u^4}{4}\right]_1^2 = \frac{1}{2} \cdot \frac{16 - 1}{4} = \frac{15}{8}$$

8. (Challenge) Find $\displaystyle\int x\sqrt{x - 1}\,dx$.

Solution

Let $u = x - 1$, so $du = dx$ and $x = u + 1$. The extra $x$ becomes $u + 1$:

$$\begin{aligned} \int (u + 1)\,u^{1/2}\,du &= \int \left(u^{3/2} + u^{1/2}\right)du \\ &= \frac{2}{5}u^{5/2} + \frac{2}{3}u^{3/2} + C \\ &= \frac{2}{5}(x - 1)^{5/2} + \frac{2}{3}(x - 1)^{3/2} + C \end{aligned}$$

9. (Challenge) Evaluate $\displaystyle\int_1^4 \frac{1}{\sqrt{x}\,(1 + \sqrt{x})^2}\,dx$.

Solution

Let $u = 1 + \sqrt{x}$, so $du = \dfrac{1}{2\sqrt{x}}\,dx$ and $\dfrac{dx}{\sqrt{x}} = 2\,du$. When $x = 1$, $u = 2$; when $x = 4$, $u = 3$.

$$2\int_2^3 u^{-2}\,du = 2\left[-\frac{1}{u}\right]_2^3 = 2\left(-\frac{1}{3} + \frac{1}{2}\right) = \frac{1}{3}$$